
\prob{00BC}{边角元Ceva定理}

\begin{figure}[htbp]
    \centering \image{00BC}
    \caption{总第~\ref{sec:00BC} 题图}
    \label{fig:00BC}
\end{figure}

如图~\ref{fig:00BC}，$P$是$\triangle ABC$内一点；作射线$AP, BP, CP$分别交$BC, CA, AB$于$P_1, P_2, P_3$。求证：
\begin{align}
  \frac{BP_1}{P_1C}\cdot\frac{CP_2}{P_2A}\cdot\frac{AP_3}{P_3B} &= 1 \label{eq:00BC-scv} \\
  \frac{\sin\angle BCP}{\sin\angle PCA}\cdot\frac{\sin\angle CAP}{\sin\angle PAB}\cdot\frac{\sin\angle ABP}{\sin\angle PBC} &= 1 \label{eq:00BC-acv}
\end{align}
\problabels{yellow/平面几何, green/证明题}

\subsection{转换比例}

\subsubsection{边元Ceva定理}

先证明式~\ref{eq:00BC-scv}。令
\[ S_1 = S_{\triangle BPC}, S_1 = S_{\triangle CPA}, S_3 = S_{\triangle APB} \]
显然有
\[ \frac{BP_1}{P_1C} = \frac{S_2}{S_3}, \frac{CP_2}{P_2A} = \frac{S_3}{S_1}, \frac{AP_3}{P_3B} = \frac{S_1}{S_2} \]
于是式~\ref{eq:00BC-scv} 显然得证。

\subsubsection{角元Ceva定理} \label{subsubsec:00BC-acv}

\begin{figure}[htbp]
  \centering \image{00BC-acv}
  \caption{步骤~\ref{subsubsec:00BC-acv} 图}
  \label{fig:00BC-acv}
\end{figure}

现在证明式~\ref{eq:00BC-acv}。如图~\ref{fig:00BC-acv}，分别作$P$在$BC, CA, AB$上的垂线，垂足分别为$H_1, H_2, H_3$。于是显然有
\begin{align*}
  \sin\angle BCP = \frac{PH_1}{PC}, & \sin\angle PCA = \frac{PH_2}{PC} \\
  \sin\angle CAP = \frac{PH_2}{PA}, & \sin\angle PAB = \frac{PH_3}{PA} \\
  \sin\angle ABP = \frac{PH_3}{PB}, & \sin\angle PBC = \frac{PH_1}{PB}
\end{align*}
于是式~\ref{eq:00BC-acv} 显然得证。

证毕。
